Why aren't the <v^2> and <vr^2> curves different in my Hernquist model?

AI Thread Summary
The discussion revolves around a problem in generating equal mass initial conditions using a Hernquist model. The user calculates the mass profile and derives velocities through a rejection method, aiming to compare the calculated <v^2> with the analytic <vr^2>. Despite expectations based on isotropy, the two curves appear identical, contradicting the theoretical relationship that <v^2> should equal 3<vr^2>. The user seeks insights into why this discrepancy does not manifest in their results. Clarification on the underlying calculations or assumptions may be necessary to resolve the issue.
LordV
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Hello, I would appreciate very much some help on this problem.

I have a Hernquist model (distribution function) and I am trying to generate equal mass initial conditions. The way I do it is this. I have the mass profile M(r) and I pick say 10^6 random numbers between 0 and 1=M(r->inf). Then I solve the inverse equation r=r(m_random) to find the radius. The next step is to draw the velocities. For each radius I calculate f(v) = f(ri,v)=f(0.5*v^2+V(r))=f(E) and then I use a rejection method to draw the velocity from f(v). Then I divide the space r (radius) in a logarithmic scale and within each bin I calculate the <v^2> from all the v's within this bin therefore I have the function <v^2>(r) and draw it and compare it with <vr^2>(r) form analytic from of the paper.

THE PROBLEM IS that the two curves are THE SAME but they shouldn't because v^2=vr^2+vphi^2+vth^2 => <v^2>=3<vr^2> because the model is isotropic i.e. <vr^2>=<vphi^2>=<vth^2>. Therefore there should be a difference by a factor of 3 in the 2 curves but there isn't any difference.

Any ideas?
 
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